Number of decrements in a dataframe

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Number of decrements in a dataframe

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#1 by (4 votes)

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I have a data frame where in each column is the value of events in the period, I need to calculate the increment and decrement of these periods for a larger base and I could not do:

p1<- rep(2:11)
p2<- rep(3:12)
p3<- rep(1:10)
p4<- rep(4:13)

base<- cbind(p1,p2,p3,p4)

base
      p1 p2 p3 p4
 [1,]  2  3  1  4
 [2,]  3  4  2  5
 [3,]  4  5  3  6
 [4,]  5  6  4  7
 [5,]  6  7  5  8
 [6,]  7  8  6  9
 [7,]  8  9  7 10
 [8,]  9 10  8 11
 [9,] 10 11  9 12
[10,] 11 12 10 13

Expected:

       p1 p2 p3 p4 in dc
  [1,]  2  3  1  4  2  1
  [2,]  3  4  2  5  2  1
  [3,]  4  5  3  6  2  1
  [4,]  5  6  4  7  2  1
  [5,]  6  7  5  8  2  1
  [6,]  7  8  6  9  2  1
  [7,]  8  9  7 10  2  1
  [8,]  9 10  8 11  2  1
  [9,] 10 11  9 12  2  1
 [10,] 11 12 10 13  2  1

That is, taking as an example line 9:

[9,] 10 11  9 12  2  1

comparing p2 with p1 there was an increase (10> 11) for period 3 there was a decrease (11 > 9) for period 4 there was another increment (9 > 12) totaling 2 increments and a decrease.

The idea is to execute this for a set of 500 variables viewed in 10 periods.

r

asked by anonymous 20.12.2017 / 17:42

1 answer

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score 4 · Accepted Answer

First I'll use the diff function to calculate the difference between two consecutive columns. It is necessary to transpose the result so that it is in the same pattern as base :

diferencas <- t(apply(base, 1, diff))
diferencas
      p2 p3 p4
 [1,]  1 -2  3
 [2,]  1 -2  3
 [3,]  1 -2  3
 [4,]  1 -2  3
 [5,]  1 -2  3
 [6,]  1 -2  3
 [7,]  1 -2  3
 [8,]  1 -2  3
 [9,]  1 -2  3
[10,]  1 -2  3

That said, I'll count how many increments there are per line. That is, I will count how many% values of% are greater than 0:

in <- apply(t(apply(diferencas, 1, function(x) x > 0)), 1, sum)

In analogous mono, I will count how many decrements per line. That is, I'm going to count how many values of diferencas are less than 0:

dc <- apply(t(apply(diferencas, 1, function(x) x < 0)), 1, sum)

Note that this works because, for diferencas , R has value 1. Now I just need to merge the results:

cbind(base, in, dc)
      p1 p2 p3 p4 in dc
 [1,]  2  3  1  4  2  1
 [2,]  3  4  2  5  2  1
 [3,]  4  5  3  6  2  1
 [4,]  5  6  4  7  2  1
 [5,]  6  7  5  8  2  1
 [6,]  7  8  6  9  2  1
 [7,]  8  9  7 10  2  1
 [8,]  9 10  8 11  2  1
 [9,] 10 11  9 12  2  1
[10,] 11 12 10 13  2  1